#1

I need the help of the UG math buffs!

Can someone help with this question Im stumped with please

Its:

A box of mass 20 KG is at rest on a rough horizontal floor. The coefficient of friction between the box and floor is 0.3. The box is subjected to horizontal force of T Newtons.

Given that T=65, find the magnitude of the acceleration of the box

Apparently the answer turns out to be 0.31 ms-2 but have not been able to get that.

Any ideas?

Can someone help with this question Im stumped with please

Its:

A box of mass 20 KG is at rest on a rough horizontal floor. The coefficient of friction between the box and floor is 0.3. The box is subjected to horizontal force of T Newtons.

Given that T=65, find the magnitude of the acceleration of the box

Apparently the answer turns out to be 0.31 ms-2 but have not been able to get that.

Any ideas?

#2

Eleventy Two.

#3

That's physics, not maths.

#4

I misread that as 'Bath Muffs'.

How strange.

How strange.

#5

nope its maths

M1 if i believe right

but yeah

Fr=(mew) N

Fr= 0.3X65

Take that away from 65 to get overall force

then F=ma

M1 if i believe right

but yeah

Fr=(mew) N

Fr= 0.3X65

Take that away from 65 to get overall force

then F=ma

#6

That's not even math, plus I suck at math half the time... lol

#7

That's physics, not maths.

i thought that as well

#8

It's maths it's mechanics

Just resolve forces and use F=ma. I'm too lazy to go through how to do it. I should be doing revision.

Just resolve forces and use F=ma. I'm too lazy to go through how to do it. I should be doing revision.

#9

None, it's at rest.

#10

That's physics, not maths.

Actually that would come under Mechanics 1 module of AS-level maths. I know cos I've just finished that module.

Use F=ma

65-.3=20a

Also, drawing a diagram usually helps me, then you can usually see how all the forces are acting on the box.

#11

First you'd find the total resistive force on the box by using F = coefficient of friction * R, so that is 0.3 * 20g which is 58.8

Then use 65-58.5 = 6.2

Then use F=ma to get a=F/m which is 6.2/20. There's your answer.

Then use 65-58.5 = 6.2

Then use F=ma to get a=F/m which is 6.2/20. There's your answer.

#12

Actually that would come under Mechanics 1 module of AS-level maths. I know cos I've just finished that module.

Use F=ma

65-.3=20a

Isn't M1 an A2 module? Meh, different boards prob.

Then Mechanics is physics

Study Physics at Uni, you'll have Pure Maths and Physics. Stuff like that would be under Physics, not pure maths.

#13

Isn't M1 an A2 module? Meh, different boards prob.

Then Mechanics is physics

Study Physics at Uni, you'll have Pure Maths and Physics. Stuff like that would be under Physics, not pure maths.

I'm doing M1 now in A2, but some schools just swap for S1 in A2,

#14

Is this kinetic or static friction? Try using F(friction)= mu*mg. this is your force in the opposite direction as your acceleration. Sum your x forces to get net force along the x-axis. then try your good old F=ma to solve for acceleration. I dont know if this is right because im too lazy to work it out right now, but try it and tell me if it works.

#15

Isn't M1 an A2 module? Meh, different boards prob.

Then Mechanics is physics

Study Physics at Uni, you'll have Pure Maths and Physics. Stuff like that would be under Physics, not pure maths.

Actually, you're right. It IS A2.

But I'm doing Maths and Further Maths at AS currently, so I generally get confused with what's what. Apart from FP1. That's definately Further maths xD

#16

gen0doom your a legend, I couldnt figure out for the hell of me how to resolve on a horizontal plane, its all those slanting planes I've resolved I tell you.

#17

I need the help of the UG math buffs!

Can someone help with this question Im stumped with please

Its:

A box of mass 20 KG is at rest on a rough horizontal floor. The coefficient of friction between the box and floor is 0.3. The box is subjected to horizontal force of T Newtons.

Given that T=65, find the magnitude of the acceleration of the box

Apparently the answer turns out to be 0.31 ms-2 but have not been able to get that.

Any ideas?

that answer is wrong actually. sig figs dictates it can only be 0.3ms-2, not 0.31.

*Last edited by muse-ik at Apr 30, 2008,*

#18

It's mechanics which is physics and math

__s__
#19

I Know! I Know! I Know!

#20

Actually, you're right. It IS A2.

But I'm doing Maths and Further Maths at AS currently, so I generally get confused with what's what. Apart from FP1. That's definately Further maths xD

Thought so.

As much as I like maths, I could never do double

#21

It's mechanics which is physics and maths

Physics is HUGELY reliant on Maths. It's not all theory, it's mostly practical...

Pure Maths isn't wordy and put into context basically.

#22

I'm doing M1 now in A2, but some schools just swap for S1 in A2,

They should. S1 is a piss take

Then again, a friend of mine who is doing double and is getting straight A's in Core and Mechanics struggles like hell at Stat, and says we should do double as we can teach it ourselves in about a month or so.

I think he's a fool, I fly through stats like anything and struggle(ish) at core.

#23

Physics is HUGELY reliant on Maths. It's not all theory, it's mostly practical...

Pure Maths isn't wordy and put into context basically.

I prefer Pure Maths. Mechanics past papers make me want to cry. Unlike Further Pure and Core modules.

Unfortunately, my class wants us to study M2 next year not S2.

#24

I prefer Pure Maths. Mechanics past papers make me want to cry. Unlike Further Pure and Core modules.

Unfortunately, my class wants us to study M2 next year not S2.

I love Stats, it's fun as hell and is so easy. Then again, it's more analytical, which is where I fly at.

#25

W = mg = 20(9.8) = 196N

Weight = Normal Force

Force of friction = Normal Force(coefficient of friction) = 196(.3) = 58.8N

T = 65

Net Force = T - Force of friction = 65 - 58.8 = 6.2N

Acceleration = Net Force / mass = 6.2 / 20 = .31m/s^2

Weight = Normal Force

Force of friction = Normal Force(coefficient of friction) = 196(.3) = 58.8N

T = 65

Net Force = T - Force of friction = 65 - 58.8 = 6.2N

Acceleration = Net Force / mass = 6.2 / 20 = .31m/s^2

#26

Physics is HUGELY reliant on Maths. It's not all theory, it's mostly practical...

Pure Maths isn't wordy and put into context basically.

its like the chicken or the egg debate? what came first physics or maths

But there are alot of similarities, and these especially show in mechanics

Physics without maths wouldn't be the same and nor would maths without physics

#27

I love Stats, it's fun as hell and is so easy. Then again, it's more analytical, which is where I fly at.

Aye, I'd take Stats over most things any day.

But, I've lost my formula book which contains my cumulative binomial distributions tables and that really sucks. Cos I need it for my homework.

#28

I love Stats, it's fun as hell and is so easy. Then again, it's more analytical, which is where I fly at.

Bah I am a bit stuck on one of my stats questions if you can shed some light on it

The question is:

The r.v. X is such that X ~ N(μ, 4). A random sample, size n , is taken from the population.

Find the least n such that P(| - μ| < 0.5) >0.95

Bearing in mind that our stats lessons got cut before christmas but it coutns for 1/3 of my a level I'm stuck.

#29

^I hate stats so much.

No.

Fr = uR

Fr=0.3x20g

65-(0.3x20g)=F

F=ma

(65-(0.3x20g))/20=a

Just figure out the numbers.

nope its maths

M1 if i believe right

but yeah

Fr=(mew) N

Fr= 0.3X65

Take that away from 65 to get overall force

then F=ma

No.

Fr = uR

Fr=0.3x20g

65-(0.3x20g)=F

F=ma

(65-(0.3x20g))/20=a

Just figure out the numbers.

#30

its like the chicken or the egg debate? what came first physics or maths

But there are alot of similarities, and these especially show in mechanics

Physics without maths wouldn't be the same and nor would maths without physics

Don't care about that

And nah, Maths has enough areas of study to keep us busy (Geometry, Statistics, Decision...) but Physics would not be Physics without Maths

#31

Bah I am a bit stuck on one of my stats questions if you can shed some light on it

The question is:

The r.v. X is such that X ~ N(μ, 4). A random sample, size n , is taken from the population.

Find the least n such that P(| - μ| < 0.5) >0.95

Bearing in mind that our stats lessons got cut before christmas but it coutns for 1/3 of my a level I'm stuck.

Haha, sorry mate, I haven't gone over normal distribution yet. I need it ages ago and I can barely remember it.

I'm doing it tomorrow, I could get back to you then if you still need help. Shame, I've revised everything but.

My stats teacher is balls, I've taught it all myself pretty much.

#32

^I hate stats so much.

I hate you so much.

Nah, you're alright.

#33

Haha, sorry mate, I haven't gone over normal distribution yet. I need it ages ago and I can barely remember it.

I'm doing it tomorrow, I could get back to you then if you still need help. Shame, I've revised everything but.

My stats teacher is balls, I've taught it all myself pretty much.

I don't even have a teacher

We only did binomial distribution then they cut the class, but its one of my maths A2 module

teach yourself the course ftl